We are interested in the limit \(c\rightarrow \infty \) , or \(\beta \rightarrow 0\) . Following [1]} we introduce \(p^2=\beta ^2 \tilde{\Delta }\)  which remains finite in this limit finite, \(p^2 \rightarrow -\tilde{\Delta }/4\tilde{c}=-h/4\) . (This is, obviously, a different quantity from the quasi-momentum \(p(\lambda )\)  mentioned above.)
We would like to find \(T\)  by summing the asymptotic expansion (REF ) while expanding it in powers of \(\beta ^2\)  which corresponds to \(1/c\)  expansion. In principle we can use the spectrum (REF ) to calculate \(\ln T\)  acting on an excited state, but resort to a simpler calculation for \(\ln T\)  acting on a primary state. In this case \(Q_{2n-1}\)  in (REF ) should be substituted by \({\sf Q}_{2n-1}^0\) , which we expand in powers of \(\beta ^2 \propto 1/c\)  (REF ). The calculation is tedious and we only give the final expression
\((\ln T)_{\rm asympt}|\Delta \rangle &=&2\pi i \sqrt{p^2-\lambda ^2} \Phi (\lambda ,p)|\Delta \rangle ,\\\nonumber \Phi (\lambda ,p)&=&1+\beta ^2\Psi -\frac{\beta ^4}{48 \lambda ^2(p^2-\lambda ^2)^3}\left[3\lambda ^4 p^2+2\pi ^2 \lambda ^4(p^2-\lambda ^2)(4p^2-3\lambda ^2)\right]\\&&\quad +\frac{\beta ^4}{2 \lambda ^2}\left[(2p^2+\lambda ^2)\Psi ^2-2(p^2-\lambda ^2)\lambda \Psi \frac{d\Psi }{d \lambda }\right]+\mathcal {O}(\beta ^6),\\\Psi (\lambda ,p)&=&\frac{\lambda ^2}{\lambda ^2-p^2}\left[\gamma +\frac{1}{2}\psi \left(2\sqrt{p^2-\lambda ^2}\right)+\frac{1}{2}\psi \left(-2\sqrt{p^2-\lambda ^2}\right)\right].\) 
